Solve Ampere's Law & Displacement Current: Capacitor Voltage Change

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The discussion focuses on calculating the rate of change of the electric field (dE/dt) between the plates of a capacitor with a diameter of 6.0 cm and a separation of 1.3 mm, given a voltage change rate of 120V/s. The area of the capacitor plates is determined to be approximately 0.006 m². The relationship between displacement current and the changing electric field is established using Ampere's Law, leading to the equation dE/dt = (μ_0I)/A. By substituting the known values, the calculated rate of change of the electric field is found to be 2.0 x 10^-4 T/s. This calculation illustrates the application of Ampere's Law in understanding displacement current in capacitors.
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Ampere's Law and Displacement Current...anybody?

Determine the rate at which the electric field changes between the round plates of a capacitor, 6.0 cm in diameter, if the plates are spaced 1.3 mm apart and the voltage across them is changing at a rate of 120V/s

I found area:

A = 2πr2 = 2 π(.03 m)2

= .006 m

then do I solve for dE/dt ?
 
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Why do you need area to find E?
 


Yes, you can solve for the rate of change of electric field (dE/dt) using the equation for displacement current, which is given by:

dE/dt = (I_d)/(Aε_0)

Where I_d is the displacement current, A is the area between the plates, and ε_0 is the permittivity of free space.

In this case, the displacement current can be calculated using Ampere's Law, which states that the line integral of the magnetic field around a closed loop is equal to the current passing through the loop. Since there is no current passing through the loop, the displacement current is equal to the rate of change of electric field.

Therefore, we can rewrite the equation as:

dE/dt = (μ_0I)/A

Where μ_0 is the permeability of free space.

Now, we can plug in the given values:

dE/dt = (μ_0(120V/s))/(.006 m)

= 2.0 x 10^-4 T/s

This is the rate at which the electric field is changing between the plates of the capacitor.
 
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